Address Translation Table Size?

Question

What will be the size of Address Translation Table, for 32b addresses if there is an entry for: a) each byte; b) each word , word size is 32b; c) each 256B unit ?

Answer 1

Word length is 32 bit  that is 4 bytes

To address a word no of entries=2^32=4GB

To address a byte =2^32*2^2=2^34=16GB

If we have 256B unit

One word 32 bit ie 4 bytes so 256B is2^8/2^2=2^6

So 64 bytes  addressed by one block

So ans for part c  will be 2^32/2^6=2^26=64MB

Comments

it is saying there is an entry for each byte. so if there is a table entry for each byte the answer is right given by pooja.

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Sorry but I am still not getting it.

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Thanks Pooja, got it. :)

Answer 2

ok. i think what they are trying to say is right. if there is a 32 bit address so it can interface 2^32 rows uniquely and now they are saying every row should also be capable of identifying byte uniquely. size of each row will be 32 bit =4 bytes now to represent 4 bytes uniquely i need 2 bits

 so answer should be  2^32*2

for first byte i will use 00, for second 01, 10 and 11 , i can't get how they multiplied it by 4.

 secondly for the second pat it should be 2^32 as every row can represent one word. i

can't get how they just did it. what i have read till now i will go with this . this is how we represent page in the page table and map .

Comments

Thank you! :)